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CTK Exchange
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Bui Quang Tuan
Member since Jun-23-07
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Jan-30-08, 01:24 PM (EST) |
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2. "RE: Two Congruent Circles From Reflections"
In response to message #1
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Thank you for your kind words, nice webpage and proof.
Of course, we can also relect C, A in BI, BX and A, B in CI, CX to get total three pairs of congruent circles.
Suppose now La is a line connected centers of two congruent circles BCiCx and CBiBx. Similarly define Lb, Lc. Three lines La, Lb, Lc bound one triangle A'B'C'. I have checked following interesting results by complicated barycentrics calculations and still not found synthetic proof:
1. From almost wellknown triangle centers (Clark Kimberling's ETC) only X = orthocenter can make ABC and A'B'C' perspective, moreover in this case they are homothetic.
2. In this case (X = orthocenter) the homothetic center is Feuerbach point of ABC and also Feuerbach point of A'B'C'.
Please note that incenter and orthocenter are closely related with Feuerbach point.
May be it is difficult to find synthetic proof for (2) but I hope that some one can find it.
Best regards,
Bui Quang Tuan
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Bui Quang Tuan
Member since Jun-23-07
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Feb-03-08, 10:28 PM (EST) |
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8. "RE: Two Congruent Circles From Reflections"
In response to message #7
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Dear Alex,
After reading D. Grinberg's paper (From Baltic Way to Feuerbach - a geometrical excursion, Mathematical Reflections 2, 2006) I think we can use theorem 10 in this paper to prove that "three tangent lines with nine point circle at A1, B1, C1 bound one triangle homothetic with ABC at Feuerbach point".
Theorem 10 states that:
"The point of tangency F of the incircle and the nine-point circle of triangle ABC is the homothetic center of the homothetic triangles A1B1C1 and XYZ"
A1 = center of circumcirles P, Q, A', Ha (your notations in latest site "Concyclic Points In A Triangle"). Similarly define B1, C1.
XYZ is intouch triangle.
Suppose three tangent lines with nine point circle at A1, B1, C1 bound one triangle A2B2C2.
From theorem 10: B1C1 // YZ and if A2 is intersection of two lines from B1, C1 and parallel with CA, AB respectively then A2B1C1 and AYZ are homothetic. By theorem 10, intersection of B1Y, C1Z is Feuerbach point therefore A2A also passes Feuerbach point. Similarly with B2B, C2C so our result is proved.
Best regards,
Bui Quang Tuan
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Bui Quang Tuan
Member since Jun-23-07
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Feb-12-08, 09:55 PM (EST) |
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10. "RE: Two Congruent Circles From Reflections"
In response to message #9
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Our Lunar New Year "Tet Festival" also takes me a break from geometry for some days. Today I see that we can prove that "three tangent lines with nine point circle at A1, B1, C1 bound one triangle homothetic with ABC at Feuerbach point" without using theorem 10. In fact, it is very easy:
Suppose three tangent lines with nine point circle at A1, B1, C1 bound one triangle T then:
- T is homothetic with ABC
- T takes nine point circle of ABC as incircle.
From this, homothetic center of ABC and T must be homothetic center of incircle of ABC and incircle of T and this center is homothetic center of incircle of ABC and nine point circle of ABC and it is Feuerbach point.There are also some easy facts when X is any point (not a orthocenter): two congruent circles from reflection (circumcircles of CBiBx and BCiCx) contain two interesting points:
1. Circumcircle of CBiBx contains incenter D of ACBx
2. Y = intersection of circumcircle of ACBx and AX (other than A), E = intersection of BBi and CY then circumcircle of CBiBx contains E.
So we have five concyclic points C, D, E, Bx, Bi. (similarly with circumcircle BCxXi). Best regards,
Bui Quang Tuan
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